If $f(x)=\left|\begin{array}{ccc}0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0\end{array}\right|$, then
(a) $f(a)=0$
(b) $f(b)=0$
(c) $f(0)=0$
(d) $f(1)=0$
Let $f(x)=\left|\begin{array}{ccc}0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0\end{array}\right|$.
Now,
$f(a)=\left|\begin{array}{ccc}0 & a-a & a-b \\ a+a & 0 & a-c \\ a+b & a+c & 0\end{array}\right|$
$=\left|\begin{array}{ccc}0 & 0 & a-b \\ 2 a & 0 & a-c \\ a+b & a+c & 0\end{array}\right|$
$=(a-b)\left(2 a^{2}+2 a c\right) \neq 0$
$f(b)=\left|\begin{array}{ccc}0 & b-a & b-b \\ b+a & 0 & b-c \\ b+b & b+c & 0\end{array}\right|$
$=\left|\begin{array}{ccc}0 & b-a & 0 \\ b+a & 0 & b-c \\ 2 a & b+c & 0\end{array}\right|$
$=(b-a)(2 a b-2 a c) \neq 0$
$f(0)=\left|\begin{array}{ccc}0 & 0-a & 0-b \\ 0+a & 0 & 0-c \\ 0+b & 0+c & 0\end{array}\right|$
$=\left|\begin{array}{ccc}0 & -a & -b \\ a & 0 & -c \\ b & c & 0\end{array}\right|$
$=a(b c)-b(a c)=0$
Hence, the correct option is (c).
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